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7 tháng 5 2016

VẾ TRÁI LÀ:

A=1/2.3+1/3.2+1/4.5+...+1/x[x+1]

A=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/n-1/n+1

A=1-1/n+1

1-1/n+1=2015/2016

1/n+1=1-2015/2016

1/n+1=1/2016

n=2016-1

n=2015

Vế trái là 

S=1/2+1/6+1/12+1/20+..+1/x(x+1)

S=1/1.2+1/2.3+1/3.4+1/4.5+...+1/x.(x+1)

S=1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+...+1/x.(x+1)

S=1-1/(x+1)=Vế phải=2015/2016

<=>1-1/(x+1)=2015/2016

1/(x+1)=1/2016

=>x+1=2016

x=2015

Ủng hộ mk mha Chí Tiến

27 tháng 4 2018

\(\frac{1}{2}\)+\(\frac{1}{6}\)+\(\frac{1}{12}\)+....+\(\frac{1}{x\left(x+1\right)}\)=\(\frac{2015}{2016}\)

\(\frac{1}{1x2}\)+\(\frac{1}{2x3}\)+\(\frac{1}{3x4}\)+.....+\(\frac{1}{x\left(x+1\right)}\)=\(\frac{2015}{2016}\)

\(1\)-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....+\(\frac{1}{x}\)-\(\frac{1}{x+1}\)=\(\frac{2015}{2016}\)

1-\(\frac{1}{x+1}\)                                                                     = \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\)                                                                         =1- \(\frac{2015}{2016}\)

\(\frac{1}{x+1}\)                                                                          = \(\frac{1}{2016}\)

\(\Rightarrow\)x + 1= 2016

\(\Rightarrow\)x = 2015

5 tháng 5 2016

\(\frac{1}{2}+\frac{1}{6}+.....+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(=\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(=1-\frac{1}{x+1}=\frac{2015}{2016}\)

\(=\frac{1}{x+1}=\frac{1}{2016}\)

=> x + 1 = 2016

=> x =2015

13 tháng 5 2016

1/2+1/6+1/12+1/20+...+1/x(x+1)=2015/2016

       1/1.2+1/2.3+1/3.4+.....+1/x.(x+1)=2015/2016

       1-1/2+1/2-1/3+1/3-1/4+......+1/x-1/x+1=2015/2016

         1-1/x-1=2015/2016

            1/x+1=1-2015/2016

               1/x+1=1/2016

 =>               x+1=2016

                 x=2016-1

                  x=2015

vậy x =2015

tích mình nha

13 tháng 5 2016

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

=>\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+.......+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

=>\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+......+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

=>\(1-\frac{1}{x+1}=\frac{2015}{2016}\)

=>\(\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)

=>x+1=2016

=>x=2015

Vậy x=2015

13 tháng 5 2016

1/2+1/6+1/12+1/20+...+1/x(x+1)=2015/2016

       1/1.2+1/2.3+1/3.4+.....+1/x.(x+1)=2015/2016

       1-1/2+1/2-1/3+1/3-1/4+......+1/x-1/x+1=2015/2016

         1-1/x-1=2015/2016

            1/x+1=1-2015/2016

               1/x+1=1/2016

 =>               x+1=2016

                 x=2016-1

                  x=2015

vậy x =2015

tích mình nha

13 tháng 5 2016

tích mình nha

7 tháng 5 2018

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2015}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(1-\frac{1}{x+1}=1-\frac{2015}{2016}\)

\(\frac{1}{x+1}=\frac{1}{2016}\)

\(x=2016-1\)

\(\Rightarrow x=2015\)

27 tháng 4 2018

1/1x2+1/2x3+...+1/x(x+1)=2015/2016

1/1-1/2+1/2-1/3+...+1/x-1/x+1=2015/2016

2/1-1/x+1=2015/2016

2016/2016-1/x+1=2015/2016

1/x+1=2016/2016-2015/2016

1/x+1=1/2016

x+1=2016

x=2016-1

x=2015

27 tháng 4 2018

         \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}.\)

 <=>\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

 <=> \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

 <=> \(-\frac{1}{x+1}=\frac{-1}{2016}\) <=>  x+1 = 2016  <=> x = 2015

a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)

\(=\dfrac{1}{2016}\)

b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)

\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)

hay x=10

Vậy: x=10

20 tháng 4 2016

\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2015}{2016}\)

\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2016}\)

\(1-\frac{1}{x+1}=\frac{2015}{2016}\)

\(\frac{1}{x+1}=1-\frac{2015}{2016}=\frac{1}{2016}\)

\(x+1=2016=>x=2015\)